# How do you find the period, amplitude and sketch y=1/100sin120pit?

May 26, 2018

The period is $= \frac{1}{60}$ and the amplitude is $= \frac{1}{100}$

#### Explanation:

We need

$\sin \left(a + b\right) = \sin a \cos b + \sin b \cos a$

The period of a periodic function is $T$ iif

$f \left(t\right) = f \left(t + T\right)$

Here,

$f \left(t\right) = \frac{1}{100} \sin \left(120 \pi t\right)$

Therefore,

$f \left(t + T\right) = \frac{1}{100} \sin \left(120 \pi \left(t + T\right)\right)$

where the period is $= T$

So,

$\frac{1}{100} \sin \left(120 \pi t\right) = \sin \left(120 \pi \left(t + T\right)\right)$

$\sin \left(120 \pi t\right) = \sin \left(120 \pi t + 120 \pi T\right)$

$\sin \left(120 \pi t\right) = \sin \left(120 \pi t\right) \cos \left(120 \pi T\right) + \cos \left(120 \pi t\right) \sin \left(120 \pi T\right)$

Then,

$\left\{\begin{matrix}\cos \left(120 \pi T\right) = 1 \\ \sin \left(120 \pi T\right) = 0\end{matrix}\right.$

$\iff$, $120 \pi T = 2 \pi$

$\iff$, $T = \frac{1}{60}$

As

$- 1 \le \sin x \le 1$

Therefore,

$- 1 \le \sin \left(120 \pi t\right) \le 1$

$- \frac{1}{100} \le \frac{1}{100} \sin \left(120 \pi t\right) \le \frac{1}{100}$

The amplitude is $= \frac{1}{100} = 0.01$

See the graph below

graph{1/100sin(120pix) [-0.0374, 0.04145, -0.02057, 0.01883]}